Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 + x}{x - 3} = \dfrac{12}{x - 3}$
Multiply both sides by $x - 3$ $ \dfrac{x^2 + x}{x - 3} (x - 3) = \dfrac{12}{x - 3} (x - 3)$ $ x^2 + x = 12$ Subtract $12$ from both sides: $ x^2 + x - (12) = 12 - (12)$ $ x^2 + x - 12 = 0$ Factor the expression: $ (x - 3)(x + 4) = 0$ Therefore $x = 3$ or $x = -4$ However, the original expression is undefined when $x = 3$. Therefore, the only solution is $x = -4$.